Proof of the identity arcsinx arccosx 2, 1 x 1

Let f(x)=arcsinx+arccosx, where f(x) is continuous at [-1,1] and derivable at (-1,1).

f'(x)=1/√(1-x^2)-1/√(1-x^2)

By the Lagrangian median value theorem.

A point can definitely be found in [-1,1].

so that f(a)=[f(1)-f(-1)] (1-(-1)).

Derived function. is equal to 0, so f(x) is a constant coefficient function i.e. f(x) = a

When x=0, f(0)=arcsin0+arccos0=2

The identity holds.

It can be done.

Prove arcsinx + arccosx = 2 arcsinx = 2-arccosx

2 sides take the sinusoid.

Left =sin(arcsinx)=x

Right =sin( 2-arccosx)=cos(arccosx)=x (using sinx=cos(2-x)).

Left = Right. That is.

Let f(x) arcsinx arccosx because f(x) is continuous in the closed interval (-1 1) and derivable in the open interval (-1 1). Because the derivative of f(x) is equal to 0

According to the Lagrangian median theorem, there is a derivative of (-1, 1) such that f(1) f(-1) 2f(c).

Because the derivative of f(c) is 0

So the f(1) f(-1) constant is 2

So f(x) is always equal to pie 2

Suppose any x=siny=cos(pi 2-y) then arcsinx=y arccosx=pi 2-y, so arcsinx+arccosx=y+pi 2-y=pi 2 is complete.

It shouldn't be f'(a)=[f(1)-f(-1)] [1-(-1)], I don't understand the following.

Let = arctan x, then cot ( 2 - = tan = x

Since 2, 2[, therefore 2 - 0,

So arccot x = 2 - i.e. arctan x + arccot x = 2

Accumulation and Difference Formula:

sinα·cosβ=(1/2)[sin(α+sin(α-

cosα·sinβ=(1/2)[sin(α+sin(α-

cosα·cosβ=(1/2)[cos(α+cos(α-

sinα·sinβ=-(1/2)[cos(α+cos(α-

and the difference product formula:

sinα+sinβ=2sin[(α/2]cos[(α/2]

sinα-sinβ=2cos[(α/2]sin[(α/2]

cosα+cosβ=2cos[(α/2]cos[(α/2]

cosα-cosβ=-2sin[(α/2]sin[(α/2]

proof of identity; arcsinx+arccosx= 2 (-1 x 1) proof: Let arcsinx = u, arccosx = v , (1 x 1), then sinu=x, cosu= [1-(sinu) 2]= 1-x 2], cosv=x, sinv = [1-(cosv) 2]= 1-x 2], left =arcsinx+arccosx=

sin(u+v)=sinuconv+conusinv=x 2+ [1-x 2] coincidental[1-x 2]=x 2+1-x 2=

1, right = sin( 2) = 1, because the trapped wide group left = right, so.

arcsinx+arccosx= 2 holds, (-1 orange x 1).

The proof is as follows:

Let f(x)=2arctanx+arcsin2x(1+x2)f'(x)

2/(1+x^2)+1/√［1-(2x/(1+x2))^2]*'

2/(1+x^2)+(1+x^2)/(1-x^2)*/1+x^2)^2

2/(1+x^2)+(1+x^2)/(1-x^2)*/1+x^2)^2=0

It can be seen that f(x)=2arctanx+arcsin2x (1+x2) is a constant function, so just enter an x value.

For example, x=1, you can get f(x)=

The identity symbol " ".

A relationship between two unraveling formulas. Given two parsed formulas, if the domain is defined for them.

, all of which have equal values, are said to be the same as these two analytic equations.

For example, x y and (x+y)(x y), for any set of real numbers (a, b), there is a b = (a+b)(a b), so x y and (x+y)(x y) are identant.

The identity of two analytic formulas cannot be discussed separately from the specified set of numbers, because the same two abysmal analytic formulas are constant in one set of numbers, and the imaginary may be non-constant in the other set of numbers. For example, x, in the set of non-negative real numbers.

Within is identant, whereas within the set of real numbers is non-contergeneous.

Let f(x)=arcsinx+arccosx:f'(x)=1 root number(1-x 2)-1 root number(1-x 2)=0

Since the derivative is equal to 0, f(x) is a constant coefficient function, i.e., f(x)=ax=0, f(0)=arcsin0+arccos0=pi 2, so the identity holds.

Let arcsin x=y,x=siny=cos(pi 2-y).

The excised limb is sold with arccos x=pi 2-y, so arcsin x+arccos x=y+pi and 2-y=pi 2

There's not one left, is there?

Proof: Rent trembling.

Let arcsinx = u, arccosx = v , (1 x 1), then sinu=x,cosu= [1-(sinu) 2]= 1-x 2],cosv=x,sinv= [1-(cosv) 2]= 1-x 2], left =arcsinx+arccosx=

sin(u+v)=sinuconv+conusinv=x^2+√[1-x^2]√[1-x^2]=x^2+1-x^2=

1, right = sin( 2) = 1, because left = right, therefore.

arcsinx+arccosx= 2 (-1 x 1).

The topic has the question of Min Qi's reputation, are you mistaken?

I've come to the conclusion that the bridge is like 2arctanx+arcsin2x (1+x 2)=

Let f(x)=arcsinx+arccosx, then it is easy to prove: renting with f'(x) 0

f (bad spring x) is constant constant.

f(0)=0+ 2

f(x) Caution 2